The components of force and velocity vectors in circular motion

[PainterGuy]

Hi!

I was trying to understand circular motion and came across two problems. I would really appreciate if you could help me with those.

Question 1:
In the picture below let's assume that the angle θ is 1 radian, i.e. 57.3°, radius is 1 m. It would mean that the length of arc AB is also 1 m. The length of chord AB is 2*r*sin(θ/2) = 2*1*sin(57.3°/2) = 1 m

How could be the length of chord and arc equal? It doesn't make any sense to me. The length of arc should be more than that of the chord. Where am I having it wrong?

Question 2:
In the picture, it might be that I'm incorrectly resolving the vectors into components. A point mass is following a circular motion in clockwise direction. The centripetal force vector Fc is resolved into Fc_x and Fc_y components, and the tangential velocity vector Vt is resolved into Vt_x and Vt_y. The components Fc_y and Vt_y are in the same direction. But weren't force vector and velocity vector supposed to not have any components parallel to each other? Where am I going wrong? Please help me.

 

[A.T.]

2*1*sin(57.3°/2) = 1 m

Really?

Fc is resolved into Fc_x and Fc_y components

The tips of Fc_x and Fc should meet.

 

[sophiecentaur]

The length of arc should be more than that of the chord.

Of course (in Euclidean geometry). You just need to continue with your derivation and take the result as the limit as θ → 0. The basic way that Calculus works, aamof.

 

[PeroK]

Question 2:
In the picture, it might be that I'm incorrectly resolving the vectors into components. A point mass is following a circular motion in clockwise direction. The centripetal force vector Fc is resolved into Fc_x and Fc_y components, and the tangential velocity vector Vt is resolved into Vt_x and Vt_y. The components Fc_y and Vt_y are in the same direction. But weren't force vector and velocity vector supposed to not have any components parallel to each other? Where am I going wrong? Please help me.

View attachment 233933

The velocity and force vectors are perpendicular. That doesn't mean that only one can have an x-component and only one a y-component.

For example, take the vectors $\vec{a} = (1, 1)$ and $\vec{b} = (1, -1)$. These are perpendicular: $\vec{a}$ is 45° above the x-axis and $\vec{b}$ is 45° below the x-axis. Yet, they both have an x-component and a y-component. And, in fact, have the same x-component.

 

[RPinPA]

The length of chord AB is 2*r*sin(θ/2) = 2*1*sin(57.3°/2) = 1 m

$\sin(57.3^\circ/2) = 0.479$, not 0.5. So chord AB is slightly less than arc AB, as you'd expect.

But weren't force vector and velocity vector supposed to not have any components parallel to each other?

Force and velocity are perpendicular. That doesn't mean they "don't have any components parallel to each other". That depends on the coordinate system you use to break into coordinates.

On a fresh diagram, draw two vectors which are perpendicular. If you choose the x-axis as the direction of one vector and the y-axis as the direction of the other, then one vector has only x-components and the other has only y-components.

But choose any other direction as an x axis. Then both vectors have an x component. And both have a y-component. There's no such rule for perpendicular vectors that they will never have parallel components in some coordinate systems.

The breaking down of a vector into perpendicular components is not unique. There are infinitely many choices of the two components. It often happens that one particular choice makes the algebra easier (for instance by making one vector purely x and the other purely y). But you could use any other choice you like.

 

[PainterGuy]

Thank you, everyone!

sin(57.3∘/2)=0.479sin⁡(57.3∘/2)=0.479\sin(57.3^\circ/2) = 0.479, not 0.5. So chord AB is slightly less than arc AB, as you'd expect.

I get it now. I was thinking that the difference between the lengths would be more, and also I was using rounded off number. Thanks.

About Question 2, I can see it now what's really going on. In the picture below, Figure 2, if point A is to follow a circular path and get to point B, its x-component, Vt_x, should gradually reduce and at the same time its y-component, Vt_y, should increase. For this to happen, Fc_x should increase and Fc_y would reduce as a consequence because Fc=Fc_x + Fc_y. At point B, you can see Vt has only y-component and Fc has only x-component.

I have another related question about elliptical orbit which I'll ask later.

Thanks a lot.

 

[A.T.]

Fc=Fc_x + Fc_y.

Yes, but this is not what your diagram shows.

 

[PainterGuy]

The tips of Fc_x and Fc should meet.

Yes, but this is not what your diagram shows.

Thank you!

But I'm sorry I don't follow you. In my first post Fc is shown as a sum of vectors Fc_y and Fc_x. The tips of Fc_x and Fc do meet. Could you please explain where I have it wrong? Thanks.

 

[haruspex]

Thank you!

But I'm sorry I don't follow you. In my first post Fc is shown as a sum of vectors Fc_y and Fc_x. The tips of Fc_x and Fc do meet. Could you please explain where I have it wrong? Thanks.

@A.T. is complaining that the tip of the black arrowhead is not at the origin. i.e. that the magnitude of the vector is to be proportional to the distance from the arrow tail to the tip of the arrowhead.
For my part, it is enough that the black line representing the vector reaches the origin; the arrowhead merely indicates the direction.

 

[A.T.]

@A.T. is complaining that the tip of the black arrowhead is not at the origin. i.e. that the magnitude of the vector is to be proportional to the distance from the arrow tail to the tip of the arrowhead.

It doesn't matter how long you draw the black Fc arrow. But it has to be consistent with the green components Fc_y, Fc_x.

 

[A.T.]

The tips of Fc_x and Fc do meet.

The tips of the black (Fc) and green (Fc_x) arrowheads must meet.

 

[PainterGuy]

Thanks.

I get your point. Those two arrows were already there then I drew other arrows myself. I agree that I should have tried to make it more consistent.